\ifx\exebook\mytoken \documentclass{exedoc}\else \fi

\docbegin
\setsection{3}

\section{Linear Operators. Fundamental Theorems}

\begin{Exec}\label{Exec4.1}
  Consider a linear functional $T : C[0, 1] \to\bc$,
  defined for every $x\in C[0, 1]$ by $Tx(t) = x(1)$.
  \begin{itemize}
  \item[(a)]  Show that $T$ is continuous in $C[0, 1]$ with respect to the
    standard norm.
  \item[(b)] Determine whether $T$ is continuous in $C[0, 1]$ with respect to
    the norm $\|x\|=\left(\int_0^1 |f(t)|^2\, dt\right)^{1/2}$, and
    justify your assertion.
  \end{itemize}
\end{Exec}

\begin{solution}
  \begin{itemize}
  \item [(a)] By the definition of the standard norm in $C[0,1]$,
    $|Tx| = |x(1)| \leq \|x\|$ for all $x\in C[0,1]$.
    Hence $T$ is bounded so that $T$ is continuous.
  \item [(b)] $T$ is not continuous in $C[0,1]$ with respect to the norm
    $\|x\|=\left(\int_0^1 |f(t)|^2\, dt\right)^{1/2}$.

    For every $\delta >0$, consider $x(t)=0$ and $y(t) = t^n$ in $C[0,1]$,
    Let $n = [2/\delta^2]+1$.
    So $\|x -y\| = \left(\int_0^1 |x(t)-y(t)|^2\, dt\right)^{1/2}
    = 1/\sqrt{2n+1} < \delta$.
    Since $|Tx - Ty| = |x(1)-y(1)| = 1$, it follows that $T$ is not continuous.
  \end{itemize}
\end{solution}

\begin{Exec}\label{Exec4.2}
  Let $h\in L^\infty[0,1]$.
  \begin{itemize}
  \item[(a)]  If $f$ is in $L^2[0,1]$, show that $fh\in L^2[0,1]$.
  \item[(b)]  Let $T: L^2[0,1]\to L^2[0,1]$ be defined as $Tf=hf$.
    Show that $T$ is a bounded linear operator.
  \end{itemize}
\end{Exec}

\begin{solution}
  \begin{itemize}
  \item [(a)] Since $\int_0^1 [f(t)h(t)]^2 dt
    \leq \int_0^1 f^2(t) \|h\|_{L^\infty[0,1]}^2 dt
    \leq \|h\|_{L^\infty[0,1]}^2 \|f\|_{L^2[0,1]}^2
    < \infty$ for every $f \in L^2[0,1]$ and $h \in L^\infty[0,1]$,
    it gives that $hf \in L^2[0,1]$.
  \item [(b)] Clearly $T$ is linear.
    By part (a), $\|Tf\|_{L^2[0,1]} = \|hf\|_{L^2[0,1]} \leq
    \|h\|_{L^\infty[0,1]} \|f\|_{L^2[0,1]}$ for all $f \in L^2[0,1]$.
    Hence $T$ is a bounded linear operator.
  \end{itemize}
\end{solution}

\begin{Exec}\label{Exec4.3}
  Suppose that a linear operator $T : X \to X$ on a normed vector space $X$
  satisfies $\|Tx\| = \|x\|$ for every $x\in X$.
  Show that $T$ is bounded, with $\|T\| = 1$.
\end{Exec}

\begin{proof}
  Since $\|Tx\| = \|x\|$ for every $x \in X$,
  $T$ is bounded and $\|T\| \leq 1$. Let $x \neq 0 \in X$.
  Then $\|T\| \geq \|Tx\|/\|x\| = \|x\|/\|x\| = 1$.
  Hence $\|T\|=1$.
\end{proof}

\begin{Exec}\label{Exec4.4}
  Consider the normed vector space $\ell^2$ of all square summate
  infinite sequences of complex numbers,
  with norm $\|x\| = \left(\sum_{i=1}^\infty |x_i|^2\right)^{1/2}$.
  For every $x = (x_1, x_2, x_3, x_4, \cdots)\in\ell^2$, let
  $$
  Tx = (0, 4x_1, x_2, 4x_3, x_4,\cdots).
  $$
  \begin{itemize}
  \item[(a)]  Show that $Tx\in\ell^2$ for every $x\in\ell^2$.
  \item[(b)]  Show that $T : \ell^2\to\ell^2$ is a bounded linear operator.
  \item[(c)]  Find the norm of $\|T\|$.
  \end{itemize}
\end{Exec}

\begin{proof}
  \begin{itemize}
  \item [(a)] Since $ \sum_{i=1}^{\infty} |4x_{2i-1}|^2 + |x_{2i}|^2 \leq
    4^2 \sum_{i=1}^{\infty} |x_i|^2 \leq 4^2 \|x\|^2 < \infty$
    for every $x \in \ell^2$,
    $\|Tx\| \leq 4 \|x\| < \infty$. So  $Tx \in \ell^2$.
  \item [(b)] Clearly $T$ is linear. By part (a), $\|Tx\| \leq 4 \|x\|$,
    so $T$ is a bounded linear operator.
  \item [(c)] Let $e_1 = (1, 0, 0, \cdots) \neq 0 \in \ell^2$.
    Since $\|T\| \geq \|Te_1\|/\|e_1\| = 4$, it gives that $\|T\| = 4$.
  \end{itemize}
\end{proof}

\begin{Exec}\label{Exec4.5}
  Suppose that $\ch$ is a Hilbert space over $\bbf$,
  and that $x_0\in \ch$ is fixed.  Show that the linear
  functional $T : \ch \to\bbf$,  defined by
  $$
  Tx = \langle x, x_0\rangle
  $$
  for every $x\in \ch$, is bounded, and find the norm of $\|T\|$.
\end{Exec}

\begin{proof}
  By the Cauch-Schwarz inequality $\|Tx\| = \inp{x}{x_0} \leq \|x\|\|x_0\|$
  for every $x \in \ch$, $T$ is bounded.
  If $\|x_0\| = 0$, $\|T\| = \sup_{x\neq 0} \|Tx\|/\|x\| = 0$.
  If $\|x_0\| \neq 0$, $\|T\| \geq \|Tx_0\|/\|x_0\| = \|x_0\|$
  so that $\|T\|=\|x_0\|$.
  Hence $\|T\| = \|x_0\|$ for every $x_0 \in \ch$.
\end{proof}

\begin{Exec}\label{Exec4.6}
  Suppose that $\ch$ is a  Hilbert space over $\bbf$,
  and that $y,z\in \ch$ are fixed.  For every $x\in \ch$, let
  $Tx =\langle x, y\rangle z$.  Show that $T : \ch\to \ch$ is a
  bounded linear operator, and find the norm of $\|T\|$.
\end{Exec}

\begin{proof}
  For evey $\alpha, \beta \in \bbf$ and $x_1, x_2 \in \ch$,
  $T(\alpha x_1 + \beta x_2) = \inp{\alpha x_1 + \beta x_2}{y}z
  = \alpha \inp{x_1}{y}z + \beta\inp{x_2}{y}z
  = \alpha Tx_1 + \beta Tx_2$, so $T$ is linear.
  Since $\|Tx\| = \|\inp{x}{y}z\| \leq \|x\|\|y\|\|z\|$, $T$ is bounded.
  Hence $T$ is a bounded linear operator.
  Clearly $\|T\|=0$ if $\|y\|=0$.
  If $\|y\|\neq 0$, $\|T\| \geq \|Ty\|/\|y\|  = \|y\|\|z\|$
  so that $\|T\| = \|y\|\|z\|$.
  Hence $\|T\| = \|y\|\|z\|$ for all $y,z\in \ch$.
\end{proof}

\begin{Exec}\label{Exec4.7}
  Show that the functions $f:C[0,1]\to\bc$ and $g : L^2[0, 1]\to \bc$,
  defined for every $f(x)=\int_0^1\sqrt{t}\, x(t^2)\,dt$ for all $x\in
  C[0,1]$ and respectively $g(x)=\int_0^1\sqrt{t}\,x(t^2)\,dt$ for all
  $x\in L^2[0, 1]$, are  bounded, with norms $\|f\|=2/3$ and $\|g\| =
  1/\sqrt{2}$. (Hint: by change of variable $s=t^2$,
  $f(x)=g(x)=\frac12\int_0^1 s^{-1/4} \, x(s)\,ds$.)
\end{Exec}

\begin{proof}
  As $x(t) \leq \|x(t)\|_\infty\ \ae{t}{[0,1]}$,
  $f(x) = \int_0^1 \sqrt{t} x(t^2) dt \leq \int_0^1 \|x\| \sqrt{t} dt
  = (2/3) \|x\|_{C[0,1]}$. So $f$ is bounded. Let $x(t) = 1$, then
  $\|f\| \geq \|f(x)\|/\|x\| = 2/3$. So $\|f\| = 2/3$

  By the Cauchy-Schwarz inequality,
  $\int_0^1 \sqrt{t} x(t^2) dt = \frac{1}{2} \int_0^1 t^{-1/4} x(t) dt
  \leq \frac{1}{2} \|t^{-1/4}\| \|x\| = \sqrt{2}^{-1} \|x\|$.
  Let $x(t) = t^{-1/4}$, then $\|g\| \geq \|g(x)\|/\|x\| = 1/\sqrt{2}$.
  So $\|g\| = 1/\sqrt{2}$.
\end{proof}

\begin{Exec}\label{Exec4.8}
  Suppose that $X$ and $Y$ are normed vector spaces over $\bbf$,
  and that $T : X\to Y$ is a bounded linear operator.
  Show that Ker\,$T = \{x \in X : Tx = 0\}$ is a closed set in $X$.
\end{Exec}

\begin{proof}
  By Theorem~\ref{Thm4.1.1} $T$ is continuous. Since $\{0\}$ is closed in $Y$,
  $\text{Ker}\,T = T^{-1}(\{0\})$ is closed in $X$ by Theorem~\ref{Thm1.6.2}.
\end{proof}

\begin{Exec}
  Suppose that $X$ and $Y$ are normed vector spaces over $\bbf$,
  and that $T : X\to Y$ is a bounded linear operator.
  \begin{itemize}
  \item[(a)]  Show that $G(T) = \{(x, Tx) : x \in X\}$
    is a linear subspace of $X\times Y$.
  \item[(b)]  Suppose that $X\times Y$  has norm defined in
    Exercise~\ref{Exec2.5} of Chapter 2.
    Show that $G(T)$ is a closed set in $X\times Y$.
    We call $G(T)$\index{Graph!of a mapping} the graph of $T$.
  \end{itemize}
\end{Exec}

\begin{proof}
  Clearly $G(T)$ is a sub set of the linear space $X\times Y$.
  Since $T$ is a linear operator and $X$ is a linear space,
  $\alpha(x, Tx)+\beta(y,Ty)
  = (\alpha x + \beta y, T(\alpha x + \beta y)) \in G(T)$ for all
  $\alpha, \beta \in \bbf$ and $x,y \in X$.

  By Theorem~\ref{Thm4.1.1} $T$ is continous.
  Let $(x,y)\in \overline{G(T)}$.
  Then there exists $x_n\in X$ such that $x_n\to x$
  and $Tx_n\to y$ as $n\to\infty$.  But $Tx_n\to Tx$ as $n\to\infty$
  and so $y=Tx$, whence $(x,y)=(x,Tx)\in G(T)$.
\end{proof}

\begin{Exec}\label{Exec4.10}
  Let $\ch$ be a complex Hilbert space and let
  $\mathcal{M}$ be a closed linear subspace of $\ch$.  If $f\in
  \mathcal{M}^\ast$, show that there exists $g\in \ch^\ast$ such that
  $g(x)=f(x)$ for all $x\in \mathcal{M}$ and $\|f\|=\|g\|$.
\end{Exec}

\begin{proof}
  Since $\mathcal{M}$ is closed linear subspace of $\ch$
  and $\ch$ is a Hilbert space, $\mathcal{M}$ is a Hilbert space.
  By the Riesz-Fr\'echet theorem, there exists a unique $y \in \mathcal{M}$
  such that $f(x)=\inp{x}{y}$ for all $x\in \mathcal{M}$ and $\|f\|=\|y\|$.
  On the other hand, $g(x)=\inp{x}{y} \in \ch^\ast$ and $\|g\|=\|y\|$.
  So $g(x)=f(x)$ for all $x\in \mathcal{M}$ and $\|f\|=\|y\|=\|g\|$.
\end{proof}

\begin{Exec}\label{Exec4.11}
  Let $c_0$ be the linear subspace of $\ell^\infty$
  consisting of all sequences which converge to $0$.
  \begin{itemize}
  \item[(a)] If $f\in (c_0)^\ast$,
    then there exists  a sequence $\{a_n\}\in\ell^1$ such that for all
    $x=\{x_n\}\in c_0$,
    \begin{equation}\label{Eq4.2.7}
      f(x)=\sum_{n=1}^\infty a_nx_n.
    \end{equation}
    Also the $(c_0)^\ast$-norm of $f$ is
    $\|f\|=\|\{a_n\}\|_1=\sum_{n=1}^\infty |a_n|$.
  \item[(b)] If $\{a_n\}\in\ell^1$ is given, then the
    right-hand side of \eqref{Eq4.2.7} defines an element of
    $(c_0)^\ast$.
  \item[(c)] Show that
    $(c_0)^\ast=\ell^1$ in the sense of isometrical
    isomorphism.
  \end{itemize}
\end{Exec}

\begin{proof}
  \begin{itemize}
  \item [(a)] Let
    $$e_n=(\underbrace{0,\cdots,0}_{n-1},1,0,\cdots)\quad\text{for all}
    \ n\in\bn.$$
    It is clear that $e_n\in c_0$ with $\|e_n\|=1$, and for each
    $x=\{\xi_n\}\in c_0$,
    we have $x=\lim\limits_{n\to\infty}\sum_{k=1}^n \xi_k\,e_k$.
    Now for each $f\in (c_0)^*$, let
    $f(e_n)=a_n$ for all $n\in\bn$, then by the continuity and
    linearity of $f$,
    $$
    f(x)=\lim_{n\to\infty}\sum_{k=1}^n \xi_k\,f(e_k)=\sum_{n=1}^\infty
    \xi_n a_n.
    $$
    Consider $x_n = (\xi_1^{(n)},\xi_2^{(n)},\xi_3^{(n)},\cdots)$ with
    $$
    \xi_k^{(n)}=\begin{cases}
      1&\text{if}\ k\leq n, a_k>=0\\
      -1&\text{if}\ k\leq n, a_k<0\\
      0&\text{if}\ k>n\ \text{or}\ a_k=0 .
    \end{cases}
    $$
    Clearly, $x_n\in c_0$.Note that
    $$
    \sum_{k=1}^n|a_k| = \sum_{k=1}^\infty \xi_k^{(n)}a_k
    =f(x_n)\leq \|f\|\,\|x_n\|
    =\|f\|\sup_{k\in \bn}|\xi_k^{(n)}|
    = \|f\|.
    $$
    Letting $n\rightarrow \infty$, we have $\sum_{k=1}^\infty|a_k| \leq \|f\|$,
    hence $\{a_k\}\in \ell^1$ and $\|\{a_k\}\|_1\leq \|f\|$.
    On the other hand,
    \[
    |f(x)|\leq \sum_{k=1}^\infty|a_k\xi_k|\leq
    \sup_{k\in\bn}|\xi_k| \sum_{k=1}^\infty|a_k|
    = \|x\|\|\{a_k\}\|_1
    \]
    Therefore $\|f\|\leq\|\{a_k\}\|_1$,
    so that $\|f\|=\|\{a_n\}\|_1=\sum_{n=1}^\infty |a_n|$.
  \item [(b)] Clearly $f$ is linear. By part~(a) $f$ is bounded.
    So $f\in (c_0)^*$.
  \item [(c)] Let $T:\ell^1\rightarrow (c_0)^\ast$
    defined by $T(a)=f$, where $f$ is defined in part (a).
    Since $T(\alpha x + \beta y)(t)
    = \sum_{k=1}^\infty (\alpha x_k + \beta y_k) t_k
    = \alpha \sum_{k=1}^\infty x_k t_k + \beta \sum_{k=1}^\infty y_k t_k
    = [\alpha T(x) + \alpha T(y)](t)$
    for every $x = \{x_k\} , y=\{y_k\}\in \ell^1, t\in c_0$
    and $\alpha, \beta\in \bc$, $T$ is linear.
    By part (a) and (b) $T$ is an isometry.
    So $(c_0)^\ast=\ell^1$ in the sense of isometrical isomorphism.
  \end{itemize}
\end{proof}

\begin{Exec}\label{Exec4.12}
  Let $c$ be the linear subspace of $\ell^\infty$
  consisting of all convergent sequences.
  \begin{itemize}
  \item[(a)] If $f\in (c)^\ast$, then there exists a number $a$
    and a sequence $\{a_n\}\in\ell^1$ such that for all
    $x=\{x_n\}\in c$,
    \begin{equation}\label{Eq4.2.8}
      f(x)=a\lim_{n\to\infty} x_n+\sum_{n=1}^\infty a_n x_n.
    \end{equation}
    Also the $c^\ast$-norm of $f$ is $\|f\|=|a|+\|\{a_n\}\|_1=|a|+\sum_{n=1}^\infty
    |a_n|$.
  \item[(b)] If $a$ and $\{a_n\}\in\ell^1$ are given, then the
    right-hand side of \eqref{Eq4.2.8} defines an element of
    $(\ell^p)^\ast$.
  \item[(c)]  $(c)^\ast=\ell^1$ in the sense of isometrical
    isomorphism.
  \end{itemize}
\end{Exec}

\begin{proof}
  \item [(a)] Since $c_0$ is a subspace of $c$, it gives that $f$ is also
    a bounded linear operator in $c_0$ so that $f\in (c_0)^\ast$.
    By Exercise~\ref{Exec4.11} then there exists
    a sequence $\{a_n\}\in \ell^1$ such that
    $f(x) = \sum_{n=1}^\infty a_n x_n$ for all $x=\{x_n\}\in c_0$.
    Note that for every $x=\{x_n\}\in c$,
    $x = b\,e + (x-b\,e)$ and $x - b\,e \in c_0$,
    where $b = \lim_{n\rightarrow \infty} x_n$ ,  $e=(1,1,1\cdots)$
    and $x - b\,e = (x_1 -b, x_2-b,x_3-b, \cdot)$.
    Letting $f(e) = \tilde{a}$, since $f$ is linear,
    $f(x) = f(b\,e + (x - b\,e)) = b\,f(e)+ f(x-b\,e)
    = b\tilde{a} + \sum_{n=1}^{\infty} a_n (x_n-b)
    = b(\tilde{a}-\sum_{n=1}^\infty a_n) + \sum_{n=1}^\infty a_n x_n
    \triangleq a \lim_{n\rightarrow \infty} x_n + \sum_{n=1}^\infty a_n x_n$.

    Consider $x_n = (\xi_1^{(n)},\xi_2^{(n)},\xi_3^{(n)},\cdots)$ with
    $$
    \xi_k^{(n)}=\begin{cases}
      1&\text{if}\ k\leq n, a_k>=0\\
      -1&\text{if}\ k\leq n, a_k<0\\
      1 &\text{if}\ k>n\ \ a>=0 \\
      -1&\text{if}\ k>n\ \ a<0.
    \end{cases}
    $$
    Note that $|f(x_n)| = |a| + \sum_{k=1}^{n} |a_k| \leq \|f\|\|x_n\| =
    \|f\|$.
    Letting $n\rightarrow\infty$, we have $|a|+\|\{a_n\}\|_1 \leq \|f\|$.
    On the other hand,
    \[
    f(x) \leq |a|\lim_{k\rightarrow \infty}x_k + \sum_{k=1}^\infty |a_k x_k|
    \leq (|a|+\sum_{k=1}^\infty |a_k|)\sup_{k\in\bn}|x_k|
    = (|a|+\|\{a_n\}\|_1)\|x\|.
    \]
    Therefore $\|f\|\leq |a|+\|\{a_n\}\|_1$,
    so that $\|f\| = |a|+\|\{a_n\}\|_1$.
  \item [(b)] Clearly $f$ is linear. By part (a) $f$ is bounded.
    So $f\in(c)^\ast$.
  \item [(c)]  Let $T:\ell^1\rightarrow (c_0)^\ast$
    defined by $T({\bf a})=f$ where $f$ is defined in part (a)  and
    ${\bf a} = (a, a_1, a_2, a_3, \infty) \in \ell^1$.
    By the same way in Exercise~\ref{Exec4.12},
    the fact that $T$ is a linear isometry follows from parts (a) and (b).
    So $(c)^\ast = \ell^1$ in the sence of isometrical isomorphism.
\end{proof}

\begin{Exec}\label{Exec4.13}
  Give examples of two norms, two linear functionals on
  $\ell^\infty$.
\end{Exec}
\begin{solution}
  The standard norm on
  $\ell^\infty$ is $\|x\| = \sup\{|x_i|:k\in\bn\}$.
  Let
  \[
  e_n=(\underbrace{0,\cdots,0}_{n-1},1,0,\cdots)\quad\text{for all} \
  n\in\bn.
  \]
  Clearly there exits a unique sequence $\{a_i\}$ such that
  $x = \sum_{i=1}^\infty a_i e_i$ for every $x\in X$.
  We will show that for every $\{b_i\}\in \ell^\infty$
  such that $b_i\neq 0$ for all $i\in \bn$,
  $\|(a_1,a_2,\cdots)\| = \sup\{|b_ia_i|:i\in\bn\}$ is a norm on $\ell^\infty$.
  In fact, $\|\{a_i^1\}+\{a_i^2\}\| =  \sup\{|b_i(a_i^1+a_i^2)|:i\in\bn\}
  \leq \sup\{|b_ia_i^1|+|b_ia_i^2)|:i\in\bn\}
  \leq \sup\{|b_ia_i^1|:i\in\bn\} + \sup\{|b_ia_i^2|:i\in\bn\}
  = \|\{a_i^1\}\|+\|\{a_i^2\}\|$,
  $\|\alpha\{a_i\}\| = \sup\{|b_i\alpha a_i|:i\in\bn\}
  = |\alpha|\sup\{|b_ia_i|:i\in\bn\} = |\alpha|\|\{a_i\}\|$,
  $\|\{a_i\}\|\geq 0$ and $\|\{a_i\}\|=0$ if and only if $|b_ia_i|=0$,
  i.e $\{a_i\} = 0$ for $b_i\neq 0$.
  So $\|\cdot\|$ is a norm on $\ell^\infty$.

  For every $n\in \bn$,
  \[
  f_n(\{a_i\}) = (\underbrace{0,\cdots,0}_{n-1},a_1,a_2,\cdots)
  \]
  is a linear functions on $\ell^\infty$.
\end{solution}

\begin{Exec}\label{Exec4.14}
  Let $\bc^N$ have the `max norm',
  $\|x\|=\max\{|x_k|: 1\leq k\leq N$, $N\geq1$ being fixed.
  Suppose that $a=(a_1, a_2,\cdots,a_N)\in\bc^N$.  Prove that
  $$
  f(x)=\sum_{k=1}^N a_k x_k
  $$
  define an element of $(\bc^N)^\ast$.  Prove that the image of the
  closed unit ball in $\bc^N$ is a closed disc in $\bc^1$.
\end{Exec}

\begin{proof}
  Clearly $f(\alpha x+\beta y) = \sum_{k=1}^N a_k(\alpha x_k+\beta y_k)
  = \alpha \sum_{k=1}^N a_k x_k + \beta \sum_{k=1}^N a_k y_k
  = \alpha f(x) + \beta f(y)$ so that $f$ is linear.
  Since
  $
  f(x) = \sum_{k=1}^N a_k x_k \leq \sum_{k=1}^N |a_k| |x_k|
  \leq \|x_k\| \sum_{k=1}^N |a_k|,
  $
  $f$ is bounded. Therefore $f\in (\bc^N)^\ast$.

  Let $B = \{x:\|x\|\leq 1\}$ is the closed unit ball in $\bc^N$.
  By the above inequation $f(x) \leq \sum_{k=1}^N |a_k|$ for every $x \in B$
  so that $f(B) \subset \{x: x\leq \sum_{k=1}^N |a_k|\} \triangleq D$.
  On the other hand, letting $x = \{x_k\}$ where $x_k = \overline{a_k}/|a_k|$
  if $a_k\neq 0$ and $x_k=0$ if $a_k=0$.
  Note that $f(zx/\sum_{k=1}^N |a_k|) = \sum_{k=1}^N z x_k a_k/ \sum_{k=1}^N |a_k|
  = z$ for every $z\in D$ and clearly $zx/\sum_{k=1}^N |a_k|\in B$.
  Hance $D \subset f(B)$ so that $D = f(B)$.
\end{proof}

\begin{Exec}\label{Exec4.15}
  Prove Corollary~\ref{Cor4.3.2}:
  If $\{T_n\}$ is a sequence of bounded linear operators each defined
  on a Banach space $X$ into a normed space $Y$, and
  $\lim_{n\to\infty} T_n(x)=T(x)$ for each $x\in X$. Then $T$ is a
  bounded linear operator on $X$ into $Y$ with
  $$
  \|T\|\leq\liminf_{n\to\infty}\|T_n\|.
  $$
\end{Exec}

\begin{proof}
  Clearly $T$ is linear follows from the linearity of $\lim$ and $T_n$.
  Since $\lim_{n\to\infty} T_n(x)$ exists for each $x\in X$,
  $\{T_n(x)\}$ is bounded for each $x\in X$. By Theorem~\ref{Thm4.3.1}
  there exits a constant $C$ such that $\|T_n\|<C$,  so that
  $\liminf_{n\to\infty}\|T_n\|<C < \infty$.

  Note that for every $\varepsilon>0$,
  there exits a integer $N$
  such that $\|T(x)\|-\|T_n(x)\|<\|T(x) - T_n(x)\|\leq \varepsilon$
  for every $n>N$.
  So $\|T(x)\|\leq \liminf_{n\to\infty} \|T_n\|\|x\|+ \varepsilon$.
  Letting $\varepsilon \rightarrow 0$, we have
  $\|T(x)\|\leq \|x\|\liminf_{n\to\infty} \|T_n\|$.
  Since $\|T\| = \sup_{x\neq 0}\|T(x)\|/\|x\|$, it follows that
  $\|T\|\leq \liminf_{n\to\infty} \|T_n\|$.
\end{proof}

\begin{Exec}\label{Exec4.16}
  If for every $x=\{x_k\}\in\ell^p (1<p<\infty)$ the series $\sum_{k=1}^\infty
  a_kx_k$ converges, prove that $a=\{a_k\}\in\ell^q$ ($1/p+1/q=1$).
\end{Exec}

\begin{proof}
  Let $T_n(x) = \sum_{i=0}^{n} a_i x_i$ and $T(x) = \sum_{i=0}^{\infty} a_i x_i$,
  then $T_n$ is a sequence of bounded linear operators
  each defined on $\ell^p$ into $\br$
  and $\lim_{n\rightarrow \infty}T_n(x) = T(x)$.
  By Corollary~\ref{Cor4.3.2} $T\in(\ell^p)^\ast$.
  Hence $\{a_n\}\in \ell^q$ follows from Theorem~\ref{Thm4.2.5}.
\end{proof}

\begin{Exec}\label{Exec4.17}
  If $\sum_{k=1}^\infty a_kx_k$ converges, whenever $x=\{x_k\}\in c$,
  prove that $a=\{a_k\}\in\ell^1$.
\end{Exec}
\begin{proof}
  Let $T_n(x) = \sum_{i=0}^{n} a_i x_i$ and $T(x) = \sum_{i=0}^{\infty} a_i x_i$,
  then $T_n$ is a sequence of bounded linear operators
  each defined on $c$ into $\br$
  and $\lim_{n\rightarrow \infty}T_n(x) = T(x)$.
  By Corollary~\ref{Cor4.3.2} $T\in(c)^\ast$.
  Hence $\{a_n\}\in \ell^1$ follows from Exercise~\ref{Exec4.12}.
\end{proof}


\begin{Exec}\label{Exec4.19}
  Prove Lemma~\ref{Lem4.4.1}:
  If $X$ is a normed linear space and $T_1, T_2\in \cb(X)$ are
  invertible then $T_1^{-1}$ and $T_2^{-1}$ are invertible with
  inverse $T_1$ and $T_2$, respectively.  Moreover $T_1T_2$ is
  invertible with inverse $T_2^{-1}T_1^{-1}$.
\end{Exec}

\begin{proof}
  Clearly $T_1T_2$ is a bijective bounded linear operator.
  Note that $T_1T_2(T_1T_2)^{-1} = I$ and $T_1,T_2$ are invertible,
  so $(T_1T_2)^{-1} = [T_2^{-1}(T_1^{-1}T_1)T_2](T_1T_2)^{-1}
  = T_2^{-1}T_1^{-1}[T_1T_2(T_1T_2)^{-1}] = T_2^{-1}T_1^{-1}$.
\end{proof}

\begin{Exec}\label{Exec4.20}
  Let $c=\{c_n\}\in\ell^\infty$ and let $T_c\in
\cb(\ell^\infty)$ be defined by $T_c(\{x_n\})=\{c_nx_n\}$.
\begin{itemize}
    \item[(a)] If
    $c_n=1/n$, show that $T_c$ is not invertible.
    \item[(b)] If $\inf\{|c_n|: n\in\bn\}>0$ and $d_n=1/c_n$, show
    that $d=\{d_n\}\in\ell^\infty$ and that $T_cT_d=T_dT_c=I$.
    \item[(c)] If $\lambda\notin \overline{\{c_n:n\in\bn\}}$, show
    that $T_c-\lambda I$ is invertible.
\end{itemize}
\end{Exec}
\begin{proof}
  \begin{itemize}
  \item [(a)] Consider $\{e_n\}$.  Since $\|T_c(e_n)\|=1/n=(1/n)\|e_n\|$
    and $1/n\rightarrow 0$ as $n\rightarrow \infty$,
    by Theorem~\ref{Thm4.4.5}, $T_c$ is not invertible.
  \item [(b)] Since $|d_n|=|1/c_n| \leq 1/\inf\{|c_n|:n\in\bn\}\leq \infty$
    for all $n\in\bn$,
    $d\in\ell^\infty$.
    Now
    \[
    T_cT_d\{x_n\} = T_c\{d_nx_n\} = \{c_nd_nx_n\} = \{x_n\}
    \]
    and
    \[
    T_dT_c\{x_n\} = T_d\{c_nx_n\} = \{d_nc_nx_n\} = \{x_n\}
    \]
    Hence $T_cT_d=T_dT_c=I$.
  \item [(c)] As $\lambda\notin \overline{\{c_n:n\in\bn\}}$
    then $\inf\{|c_n - \lambda|:n\in \bn\}>0$.
    Hence if $b_n=c_n-\lambda$ then $\{b_n\}\in \ell^\infty$
    and as $\inf\{b_n:n\in\bn\}>0$ then $T_b$ is invertible by part~(b).
    But $T_b\{x_n\} = \{(c_n-\lambda)x_n\} = \{c_nx_n\} - \{\lambda x_n\}
    = (T_c -\lambda I)\{x_n\}$ so $T_b = T_c -\lambda I$.
    Thus $T_c-\lambda I$ is invertible.
  \end{itemize}
\end{proof}

\begin{Exec}\label{Exec4.21}
  Let $(X,\|\cdot\|_1)$, $(X,\|\cdot\|_2)$ be Banach spaces such that
  $\|x\|_1\leq k \|x\|_2$ for all $x\in X$ and some constant $k>0$.
  Show that $\|\cdot\|_1$ and $\|\cdot\|_2$ are equivalent.
\end{Exec}
\begin{proof}
  The identity map $I:X\rightarrow X$ from the Banach space
  $(X,\|\cdot\|_2)$ to $(X,\|\cdot\|_1)$ is a bounded linear operator.
  By Theorem~\ref{Thm4.4.5} since $(1/k) \|x_1\|_1 < \|x\|_2 = \|I(x)\|_2$,
  $I$ is invertible. Hence there exits $K>0$ such that $\|I(x)\|_2 < K\|x\|_1$
  so that $\|\cdot\|_1$ and $\|\cdot\|_2$ are equivalent.
\end{proof}

\begin{Exec}\label{Exec4.22}
  Let $T_1\in \cb(X_1,X_3)$, $T_2\in \cb(X_2,X_3)$, where
  $X_1,X_2,X_3$ are Banach spaces.  If the equation $T_1(x)=T_2(y)$ has
  for every $x$ a unique solution $y=T(x)$, prove that $T$ is a
  bounded linear operator on $X_1$ into $X_2$. (Hint: use the closed
  graph theorem)
\end{Exec}
\begin{proof}
  Since $G(T_1)$ and $G(T_2)$ are closed follow from the closed graph theorem.
  Let $G(T)=\{(x,y):(x,T_1(x))\in G(T_1), (y,T_2(y))\in G(T_2), T_1(x)=T_2(y)\}$.
  Since $T_1(x)=T_2(y)$ has a unique solution for every $x$,
  $G(T)$ is the graph of $T$. For every sequence $\{(x_n,y_n)\}\subset G(T)$
  such that $(x_n,y_n) \rightarrow (x,y)$,
  by the continuous of $T_1$ and $T_2$, $(x_n,T_1(x_n))\rightarrow(x,T_1(x))$,
  $(y_n,T_2(y_n))\rightarrow(y,T_2(y))$ and $T_1(x)=T_2(y)$.
  Hence $(x,y)\in G(T)$ so that $G(T)$ is closed.
  $T\in \cb(X_1,X_2)$ follows from the closed graph theorem.
\end{proof}

\begin{Exec}\label{Exec4.23}
  Let $X, Y$ be Banach spaces and let the operator $T_0\in
  \cb(X,Y)$ be bijective.  If the operator $T_1\in \cb(X,Y)$ is such
  that $\|T_0^{-1}\|\, \|T_1\|<1$, prove that the operator $T_0+T_1$
  is invertible. (Hint: use the Banach contraction principle of
  Chapter~1)
\end{Exec}

\begin{proof}
  For every $y\in Y$, consider $f(x) = -T_0^{-1}T_1 x+T_0^{-1}y$.
  Since $\|f(x_1)-f(x_2)\| = \|T_0^{-1}T_1(x_2-x_1)\|
  \leq\|T_0^{-1}\|\|T_1\|\|x_1-x_2\|$ and $\|T_0^{-1}\|\|T_1\|<1$,
  $f$ has a unique fixed point by the Banach fixed point principle.
  Since $T_0$ is bijective, $T_0x = T_0f(x) = -T_0 T_0^{-1}T_1 x + T_0T_0^{-1}y$
  has a unique solution $x\in X$ so that $y = (T_0+T_1)x$ has a unique solution.
  Therefore $T_0+T_1$ is bijective. Clearly $T_0+T_1\in \cb(X,Y)$.
  Hence $T_0+T_1$ is invertible by the Banach theorem.
\end{proof}

\begin{Exec}\label{Exec4.24}
  Use Theorem~\ref{Thm4.4.5} to reconsider
  Example~\ref{Exam4.4.4}.
\end{Exec}

\begin{proof}
  As
  $\|T(f)\| = \left(\int_0^1 |(t+1)f(t)|^2\right)^{1/2}
  \geq \left(\int_0^1 |f(t)|^2\right)^{1/2}$ for all $x\in L^2[0,1]$,
  $T$ is invertible by Theorem~\ref{Thm4.4.5}.
\end{proof}

\begin{Exec}
  Let $X$ be a normed linear space.  Suppose that
  $x_1,x_2,\cdots,x_k$ are $k$ linearly independent vectors in $X$,
  $\alpha_1,\alpha_2,\cdots, \alpha_k$ are numbers in $\bbf$.  Prove
  that there exists an $f\in X^\ast$ with $f(x_\nu)=\alpha_\nu$
  ($\nu=1,2,\cdots,k$) and $\|f\|\leq M$ if and only if for all
  $t_1,t_2,\cdots,t_k\in \bbf$
  $$
  \left|\sum_{\nu=1}^k t_\nu \alpha_\nu \right|\leq
  M\left\|\sum_{\nu=1}^k t_\nu x_\nu\right\|
  $$
  holds.
\end{Exec}

\begin{proof}
  Clearly if $f\in X^\ast$ with $f(x_\nu)=\alpha_\nu$
  ($\nu=1,2,\cdots,k$) and $\|f\|\leq M$,
  for all $t_1,t_2,\cdots,t_k\in \bbf$
  $$
  \left|\sum_{\nu=1}^k t_\nu \alpha_\nu \right| =
  |f(\sum_{\nu=1}^k t_\nu x_\nu)|
  \leq \|f\|\left\|\sum_{\nu=1}^k t_\nu x_\nu\right\|
  \leq M\left\|\sum_{\nu=1}^k t_\nu x_\nu\right\|
  $$

  On the other hand,
  for every $\sum_{\nu=1}^k t_\nu x_\nu \in \sspan\{x_k:k\in\bn\}=M$,
  let $\tilde{f}(\sum_{\nu=1}^k t_\nu x_\nu) = \sum_{\nu=1}^k t_\nu \alpha_\nu$.
  So $\tilde{f}$ is a linear and $|\tilde{f}(x)|\leq M\|x\|$ on $M$.
  By Hahn-Banach theorem, there exists $f$ such  that $f(x)\leq M\|x\|$ on $X$
  and $f(x) = \tilde{f}(x)$ on $M$,
  i.e $f(x_\nu) = \tilde{f}(x_\nu) = \alpha_\nu$.
  Putting $-x$ in place of $x$ we deduce that $|f(x)|\leq M\|x\|$.
  Hence $\|f\|\leq M$ and $f\in X^\ast$.
\end{proof}

\begin{Exec}\label{Exec4.26}
  Let $X$ be a normed linear space and let $Z$ be a linear
  subspace of $X$.  Suppose that $x_0\in X$ such that $d(x_0,
  Z)=\inf_{z\in Z}d(x_0,z)>0$.  Prove that there exists an $f\in
  X^\ast$ such that
  \begin{itemize}
  \item[(1)]  $f(z)=0$ whenever $z\in Z$;
  \item[(2)]  $f(x_0)=d(x_0,Z)$;
  \item[(3)]  $\|f\|=1$.
  \end{itemize}
\end{Exec}

\begin{proof}
  Let $M = \{x_0\}+Z = \{\alpha x_0+z: z\in Z\}$.
  Clearly $M$ is a linear subspace of $X$.
  We claim that there exists a unique $\alpha\in\bbf$ such that
  $x = \alpha x_0 + z_0$ and $z_0\in Z$ for every $x\in M$.
  In fact if there exist $\alpha_1, \alpha_2 \in \bbf$ such that
  $x = \alpha_1 x_0 + z_1$ and $x = \alpha_2 x_0 + z_2$,
  it gives $(\alpha_1-\alpha_2)x_0 = z_2 - z_2 \in Z$.
  Since $d(x,Z) >0$ and $Z$ is a linear subspace, $\alpha_1=\alpha_2$.

  Let $f(x) = f(\alpha x_0+z_0) = \alpha d(x_0,Z)$ for every $x\in M$.
  Clearly $f(z) = 0$ for $z\in Z$, $f(x_0)= d(x_0,Z)$ and $f$ is linear in $M$.
  Since $|f(x)| = |\alpha| d(x_0,Z)
  = |\alpha|\inf_{z\in Z}\|x_0-z\|
  = \inf_{z\in Z} \|\alpha x_0 - \alpha z\|
  = \inf_{z\in Z} \|\alpha x_0 + z\|
  \leq \|\alpha x_0 + z_0\| = \|x\|$, $f$ is  bounded.
  For $\inf_{z\in Z} \|x_0-z\|$ exists, there exists a sequence $\{z_n\}\in Z$
  such that $ d(x_0,z_n) = \|x_0-z_n\| \rightarrow d(x_0,Z)$.
  Hence $|f(x_0-z_n)|/\|x_0-z_n\| = d(x_0,Z) / \|x_0-z_n\| \rightarrow 1$
  so that $\|f\|=1$.

  By Corollary~\ref{Cor4.5.1} $f$ can be extended linearly
  with preservation of the norm to the whole of $X$.

\end{proof}

\begin{Exec}\label{Exec4.27p}
  Prove that the dual space of a infinite-dimensional
  normed linear space is infinite-dimensional.
\end{Exec}

\begin{proof}
  Suppose that $X$ is an infinite-dimensional normed linear space.
  If $X^\ast$ is finite-dimensional, $X^{\ast\ast}$ is finite-dimensional
  (isometrically isomorphic to $Mat(n\times n)$,
  where $n$ is the dimension of $X^\ast$).
  On the other hand, by Theorem~\ref{Thm4.5.4} $X$ is isometrically isomorphic
  to a subspace $\tilde{X}$ of the bidual $X^{\ast\ast}$ so that
  $X^{\ast\ast}$ is infinite-dimensional, a contradiction.
  Hence $X^\ast$ is infinite-dimensional.
\end{proof}

\begin{Exec}\label{Exec4.27}
  Let $X$ is a finite-dimensional normed linear space.
  Prove that the strong convergence and the weak convergence of a
  sequence in $X$ are equivalent.
\end{Exec}

\begin{proof}
  Since $X$ is a finite-dimensional space,
  there exits a basis $\{e_i\}_{i=1,\cdots,n}$ of $X$.  Hence there
  exits a unique sequence $\{a_i\}_{i=1,\cdots,n}\in \bbf^n$ such that
  $x = \sum_{i=1}^{n} a_i e_i$ for every $x\in X$.
  Suppose that  $\{x_k = \sum_{i=1}^{n} a_i^k e_i\}_{k\in\bn}$
  converges weakly to $x = \sum_{i=1}^{n} a_i e_i\}$,
  let $f_i(x) = f_i(\sum_{i=1}^{n} a_i e_i) = a_i$. Clearly $f_i\in X^\ast$.
  As $|f_i(x_k) - f_i(x)| = |f_i(x_k-x)| = |a_i^k - a_i| \rightarrow 0$
  for all $i = 1,\cdots, n$,
  $\|x_k - x\| = \| \sum_{i=1}^{n} (a_i^k-a_i) e_i\|
  \leq \sum_{i=1}^{n} \|a_i^k-a_i\|\|e_i\| \rightarrow 0$.
  So $x_k \rightarrow x$ as $k\rightarrow \infty$, this, together with
  Remark~\ref{Rek4.6.1}, implies the equivalence of
  the strong convergence and the weak convergence.
\end{proof}



\begin{Exec}\label{Exec4.28}
  Let $M$ be a closed subspace of a normed linear space $X$.  Suppose
  that a sequence $\{x_n\}\subset M$ converges weakly to $x_0$.  Show
  that $x_0\in M$. (Hint: using Exercise~4.26)
\end{Exec}

\begin{proof}
  Suppose that $\{x_n\}\subset M$ converges weakly to $x_0$.
  If $x_0\notin M$, $d(x_0,M)>0$ as $M$ is closed.
  By Exercise~\ref{Exec4.26} there exits $f\in X^*$ such that
  $f(x_0) = d(x_0,M)$ and $f(z) = 0$ when $z\in M$.
  Since $x_n\in M$,  $f(x_n)\rightarrow 0 \neq f(x_0)$, a contradiction.
  Hence $x_0 \in M$.
\end{proof}

\begin{Exec}\label{Exec4.29}
   Let $\ch$ be a Hilbert space.
   Suppose that $x_0$, $x_n\in \ch$ ($n=1,2,\cdots$).
   Prove that if $\{x_n\}$ converges weakly to $x_0$ ($n\to\infty$),
   and $\|x_n\|\to \|x_0\|$ ($n\to\infty$),
   then $\{x_n\}$ converges strongly to $x_0$.
\end{Exec}

\begin{proof}
  Since $\{x_n\}$ converges weakly to $x_0$, letting $f(x) =\inp{x}{x_0}$,
  $f\in \ch^\ast$ so that $|f(x_n)-f(x_0)| = |\inp{x_n-x_0}{x_0}|
  \rightarrow 0$.
  As $\|x_n\|\rightarrow \|x_0\|$,
  $\|x_n-x_0\| = \inp{x_n-x_0}{x_n-x_0}
  = -\inp{x_n-x_0}{x_0} + \inp{x_n}{x_n} - \inp{x_0}{x_0} + \inp{x_0}{x_0-x_n}
  \rightarrow 0$.
  Hence $\{x_n\}$ converges strongly to $x_0$.
\end{proof}

\begin{Exec}\label{Exec4.30}
  Let $X$ be a separable Banach space and let $M$ be a
  bounded subset of $X^\ast$.  Prove that each sequence of $M$
  contains a weakly$^\ast$ convergent subsequence.
\end{Exec}

\begin{proof}
  Since $X$ is separable, there exists
  a countable dense subset $D=\{x_i\}_{i\in\bn}$ of $X$.
  Suppose that $\{f_i\}_{i\in\bn}$ is a sequence of $M$.
  Since $M$ is bounded and $|f_i(x_0)|\leq \|f_i\|\|x_0\|$,
  $\{f_i(x_0)\}$ is bounded in $\bbf$ so that
  there exits a subsequence $\{f_{n_i^0}(x_0)\}_{i\in\bn}$ which converges.
  For each $k>0$, clearly $\{f_{n_i^{k-1}}(x_k)\}$ is bounded,
  so there exits a subsequence $\{f_{n_i^k}(x_k)\}_{i\in\bn}$ wich converges.
  Considering $\{f_{n_k^k}\}$, $\{f_{n_k^k}(x)$ converges for every $x\in D$ by
  the definition of $\{f_{n_k^k}\}$.
  Clearly $\{f_{n_k^k}\}$ is bounded. Hence $\{f_{n_k^k}\}$ is a
  weakly$^\ast$ convergent subsequence by Theorem~\ref{Thm4.6.2}
  (Letting $Y=\bbf$, then strongly convergent means wealky$^\ast$ convergent).
\end{proof}

\begin{Exec}\label{Exec4.31}
  Let $X$ be a Banach space and let $T:X\to X$ is a
  linear operator.  Prove that $T$ is bounded if and only if  for
  every $x_n\rightharpoonup x$ ($n\to\infty$), we have $Tx_n\rightharpoonup
  Tx$ as $n\to\infty$. (Hint: use of the closed graph theorem)
\end{Exec}

\begin{proof}
  If $T$ is bounded, $fT(x)$ is bounded for every $f\in X^\ast$.
  So $f(Tx_n)\rightarrow f(Tx)$ for every $f\in X^\ast$ if
  $x_n \rightharpoonup x$.

  On the other hand, suppose that for
  every $x_n\rightharpoonup x$ ($n\to\infty$), we have $Tx_n\rightharpoonup
  Tx$ as $n\to\infty$.
  For all sequence $\{(x_n, Tx_n)\}$ in $G(T)$
  convergences to $(x,y)\in G(T)$, $x_n \rightharpoonup x$ and
  $Tx_n \rightharpoonup y$ by Remark~\ref{Rek4.6.1} so that
  $|f(Tx-y)| \leq  |f(Tx_n)-f(Tx)| + |f(y)-f(Tx_n)|\rightarrow 0$, i.e
  $|f(Tx-y)| = 0$
  for all $f\in X^\ast$.
  So $Tx = y$ by Corollary~\ref{Cor4.5.3}, it gives that $G(T)$ is closed.
  By the closed graph theorem, $T$ is bounded.
\end{proof}

%%\begin{Exec}\label{Exec4.32}
%%  Let $\{x_n\}$ be a sequence of $C[a,b]$ and $x_0\in C[a,b]$.  Prove that
%%  $\{x_n\}$ converges weakly to $x_0$ if and only if there exists a
%%  constant $M>0$ such that $\|x_n\|\leq M$ for all $n\in\bn$, and
%%  $\lim_{n\to\infty} x_n(t)=x_0(t)$ for all $t\in [a,b]$.
%%\end{Exec}


\docend
